# Thread: Indeterminant limit involving an integral.

The value of is :
(a) 0
(b)1/12
(c)1/24
(d)1/64

First, expand to series t(1+t)/(4+t^4) = t²/4 - (t^3)/8 +O(t^4)
Second, integration from t=0 to x leads to I= (x^3)/12 - (x^4)/32 +O(x^5)
Then, I/x^3 = 1/12 + x/32 +O(x²) which limit for x -> 0 is 1/12

Originally Posted by sbhatnagar
The value of is :
(a) 0
(b)1/12
(c)1/24
(d)1/64
Use l'Hospital's Rule:

The derivative of the numerator is (using the Fundamental Theorem of Calculs) $\frac{x \ln(1 + x)}{x^4 + 4}$ and the derivative of the denominator is 3x^2. Therefore ....

$\lim_{x\to \0 }\frac{1}{x^3}\int_{0}^{x}\frac{tln(t+1)}{4+t^4}dt$
$=\lim_{x\to \0 }\frac{1}{3x^2}\frac{xln(x+1)}{4+x^4}$
$=\lim_{x\to \0 }\frac{ln(x+1)}{3x(4+x^4)}$
$=\frac{1}{12}$